What is the integral of tan^3(2x)sec^4(2x) ?

The integral of tan^3(2x)sec^4(2x) is (sec^6(2x)}{12}-\frac{sec^4(2x))/8+C

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integral of tan^3(2x)sec^4(2x)

\int tan^{3} (2 x) sec^{4} (2 x) d x

\frac{sec ^{6} ( 2 x )}{12} - \frac{sec ^{4} ( 2 x )}{8} + C

Solution steps

\int tan^{3} (2 x) sec^{4} (2 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (2 x)) tan (2 x) sec^{4} (2 x) d x

Apply u-substitution

: \int \frac{u ^{3} ( u ^{2} - 1 )}{2} d u

= \int \frac{u ^{3} ( u ^{2} - 1 )}{2} d u

Expand

\frac{u ^{3} ( u ^{2} - 1 )}{2} : \frac{u ^{5}}{2} - \frac{u ^{3}}{2}

= \int \frac{u ^{5}}{2} - \frac{u ^{3}}{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u ^{5}}{2} d u - \int \frac{u ^{3}}{2} d u

\int \frac{u ^{5}}{2} d u = \frac{u ^{6}}{12}

\int \frac{u ^{3}}{2} d u = \frac{u ^{4}}{8}

= \frac{u ^{6}}{12} - \frac{u ^{4}}{8}

Substitute back

u = sec (2 x)

= \frac{sec ^{6} ( 2 x )}{12} - \frac{sec ^{4} ( 2 x )}{8}

Add a constant to the solution

= \frac{sec ^{6} ( 2 x )}{12} - \frac{sec ^{4} ( 2 x )}{8} + C

\int_{x}^{x + 2} (4 t + 1) d t

\frac{\partial}{\partial x} (2 cos (x y))

y = 2 x^{3} - 5 x, (1, - 3)

f (x) = 2 x^{3}, (- 1, - 2)

\frac{\partial}{\partial x} (\frac{1}{2} sin (2 x - y))

What is the integral of tan^3(2x)sec^4(2x) ?
The integral of tan^3(2x)sec^4(2x) is (sec^6(2x)}{12}-\frac{sec^4(2x))/8+C